Browserstack Placement Papers 2026
Last Updated: March 2026
Company Overview
BrowserStack is a cloud web and mobile testing platform enabling developers to test applications across 3000+ real devices and browsers.
Key Facts:
- Founded: 2011
- Employees: 1000+
- India Offices: Mumbai (HQ)
- Valuation: $4 billion
Eligibility Criteria
| Criteria | Requirement |
|---|---|
| Degree | B.Tech/B.E, M.Tech, MCA |
| Branches | CSE, IT |
| Academic Score | 70%+ throughout |
| Backlogs | No active backlogs |
CTC & Compensation
| Role | CTC (Fresher) |
|---|---|
| Software Engineer | 10-18 LPA |
| QA Engineer | 8-14 LPA |
| DevOps Engineer | 10-16 LPA |
Exam Pattern
| Section | Questions | Duration |
|---|---|---|
| Aptitude | 15 | 20 min |
| Technical MCQ | 20 | 25 min |
| Coding | 3 | 90 min |
Aptitude Questions
Q1
Find next: 1, 4, 9, 16, 25, ? Answer: 36 (squares)
Q2
20% of A = 30% of B. If A=150, find B. Answer: 100
Q3
Train 120m long at 72 km/hr crosses bridge in 15 sec. Bridge length? Answer: 180m
Q4
Ages ratio 2:3, sum 50. After 10 years ratio? Answer: 3:4
Q5
CP of 50 articles = SP of 40 articles. Profit %? Answer: 25%
Technical Questions
- Difference between Selenium and Cypress?
- What is CI/CD pipeline?
- Docker vs Virtual Machines?
- Test automation frameworks?
- Cross-browser testing challenges?
Coding Questions (Python)
Q1: Implement Queue using Stacks
class QueueUsingStacks:
def __init__(self):
self.stack1 = []
self.stack2 = []
def enqueue(self, x):
self.stack1.append(x)
def dequeue(self):
if not self.stack2:
while self.stack1:
self.stack2.append(self.stack1.pop())
return self.stack2.pop() if self.stack2 else None
Q2: First Non-Repeating Character
def first_non_repeating(s):
from collections import Counter
count = Counter(s)
for char in s:
if count[char] == 1:
return char
return None
Q3: Trie Implementation
class TrieNode:
def __init__(self):
self.children = {}
self.is_end = False
class Trie:
def __init__(self):
self.root = TrieNode()
def insert(self, word):
node = self.root
for char in word:
if char not in node.children:
node.children[char] = TrieNode()
node = node.children[char]
node.is_end = True
def search(self, word):
node = self.root
for char in word:
if char not in node.children:
return False
node = node.children[char]
return node.is_end
Interview Tips
- Testing Knowledge: QA mindset and practices
- Automation: Understanding of test automation
- DevOps Basics: CI/CD, Docker basics
- Problem Solving: Algorithmic thinking
- Product Understanding: Know BrowserStack products
FAQs
Q1: Is BrowserStack good for freshers? Yes, product-focused culture Q2: Remote work? Remote-first culture Q3: Tech focus? Testing, Cloud, DevOps
All the best for your BrowserStack placement!
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